Noesis
Interleaved — Phase 5

Biology interleaved practice

printable — key stays hidden on paper

Instructions: Answer all problems. Each requires you to identify the correct concept or method before solving. Show working for calculations. Marks shown in brackets. Use ...... for any math.

  1. A meadow contains 240 grasshoppers spread across 0.6 km20.6\ \text{km}^2. Calculate the population density in grasshoppers per km2\text{km}^2, and state whether density alone tells you the distribution pattern. [3]

  2. In a food chain, producers capture 50,000 kJ m−2 yr−150{,}000\ \text{kJ}\,\text{m}^{-2}\,\text{yr}^{-1} of energy. Using the 10% rule, how much energy is available to a secondary consumer? Show each step. [3]

  3. Classify each interaction as mutualism, commensalism, or parasitism: (a) tapeworm in a human gut; (b) barnacles attached to a whale; (c) mycorrhizal fungi on plant roots exchanging nutrients for sugars. [3]

  4. A newly formed volcanic island is colonized by lichens, then mosses, then shrubs. Name this type of succession and explain the one key feature that distinguishes it from the other type. [3]

  5. Distinguish r-selected from K-selected species using two contrasting traits, and match each to the survivorship curve (Type I, II, or III) it most likely follows. [4]

  6. Identify each factor as biotic or abiotic, then say whether it acts as density-dependent or density-independent on a population: (a) a disease outbreak; (b) a sudden frost; (c) competition for food. [4]

  7. Name the process in the nitrogen cycle that converts atmospheric N2\text{N}_2 into a usable form, name the organisms responsible, and name the process that returns N2\text{N}_2 to the atmosphere. [3]

  8. A pond ecosystem loses its top predator (a keystone species). Predict two ecosystem-level consequences and explain the mechanism behind one of them. [3]

  9. A population grows according to the logistic model with carrying capacity K=800K = 800 and current size N=600N = 600. If the maximum per-capita growth rate is rmax=0.10r_{max}=0.10, calculate the actual population growth rate dNdt\dfrac{dN}{dt} at this moment using dNdt=rmaxN(K−NK)\dfrac{dN}{dt}=r_{max}N\left(\dfrac{K-N}{K}\right). [4]

  10. For a grassland, an ecological pyramid of numbers can appear inverted (a single tree supporting many insects), yet the energy pyramid never inverts. Explain why the energy pyramid is always upright, and briefly relate this to the carbon cycle's role in producers. [4]

Answer keyMark scheme & solutions

Q1 — Subtopic 5.2.1 (population density & distribution). Density =2400.6=400 grasshoppers/km2= \dfrac{240}{0.6} = 400\ \text{grasshoppers/km}^2. Density alone does not reveal distribution — the same density could be clumped, uniform, or random. Distribution describes spatial arrangement, a separate measure. [3] Why: Numbers cue a density calculation, but part two tests the conceptual distinction (density ≠ dispersion pattern).

Q2 — Subtopic 5.1.7 (10% rule). Producer → primary consumer: 50,000×0.10=5,000 kJ50{,}000 \times 0.10 = 5{,}000\ \text{kJ}. Primary → secondary consumer: 5,000×0.10=500 kJ5{,}000 \times 0.10 = 500\ \text{kJ}. Secondary consumer gets 500 kJ m−2 yr−1500\ \text{kJ}\,\text{m}^{-2}\,\text{yr}^{-1}. [3] Why: Requires two applications of the 10% rule (two trophic steps), not one — students must count trophic levels correctly.

Q3 — Subtopic 5.2.9 (symbiosis). (a) Parasitism (tapeworm benefits, host harmed). (b) Commensalism (barnacle benefits, whale unaffected). (c) Mutualism (both benefit). [3] Why: Forces distinguishing three symbiosis types by who gains/loses.

Q4 — Subtopic 5.1.11 (succession). This is primary succession. Key distinguishing feature: it begins on bare rock/lifeless substrate with no pre-existing soil, so pioneer species (lichens) must first form soil. Secondary succession occurs where soil already exists after a disturbance. [3] Why: Both types share the sequence idea; the discriminating detail is presence/absence of soil.

Q5 — Subtopic 5.2.5 + 5.2.6 (r/K-selection + survivorship). r-selected: many offspring, little parental care, small body, early maturity → Type III curve (high early mortality). K-selected: few offspring, high parental care, large body, late maturity → Type I curve (high survival until old age). [4] Why: Interleaves two subtopics — students must connect life-history strategy to survivorship shape.

Q6 — Subtopics 5.1.2 + 5.2.4 (biotic/abiotic + density factors). (a) Disease = biotic; density-dependent. (b) Frost = abiotic; density-independent. (c) Competition = biotic; density-dependent. [4] Why: Two independent classifications per item prevents pattern-matching a single answer.

Q7 — Subtopic 5.1.9 (nitrogen cycle). Nitrogen fixation converts N2\text{N}_2 to ammonia/ammonium; carried out by nitrogen-fixing bacteria (e.g., Rhizobium, cyanobacteria). Denitrification (by denitrifying bacteria) returns N2\text{N}_2 to the atmosphere. [3] Why: Tests recall of specific named steps, easily confused with nitrification.

Q8 — Subtopic 5.2.10 (keystone species). Consequences: (1) prey population booms; (2) reduced biodiversity as the dominant prey outcompetes others. Mechanism: without predation controlling the dominant competitor, that species monopolizes resources, driving local extinctions — a trophic cascade. [3] Why: Requires cause-and-effect reasoning specific to keystone (not just any) species.

Q9 — Subtopics 5.2.2/5.2.3 (logistic growth + carrying capacity). dNdt=0.10×600×800−600800=0.10×600×0.25=15\dfrac{dN}{dt}=0.10 \times 600 \times \dfrac{800-600}{800} = 0.10 \times 600 \times 0.25 = 15 individuals per unit time. [4] Why: Students must plug into the logistic (not exponential) equation — recognizing KK appears is the method choice.

Q10 — Subtopics 5.1.5/5.1.6 + 5.1.8 (energy flow/pyramids + carbon cycle). Energy pyramids never invert because energy is lost at each transfer (only ~10% passes up; rest lost as heat/respiration), so each level always has less energy than the one below. Number/biomass pyramids can invert because they count individuals/mass, not energy. Producers fix atmospheric carbon (CO2_2) via photosynthesis, forming the energy-rich base that feeds all higher levels. [4] Why: Integrates the second law logic of energy flow with the carbon cycle's input at the pyramid base.

[
  {"claim":"Q2: secondary consumer receives 500 kJ after two 10% steps from 50000 kJ","code":"E=50000\nfor _ in range(2):\n    E*=0.10\nresult=(E==500)"},
  {"claim":"Q9: logistic dN/dt equals 15 for r=0.10, N=600, K=800","code":"r=0.10\nN=600\nK=800\ndNdt=r*N*(K-N)/K\nresult=(dNdt==15.0)"},
  {"claim":"Q1: density is 400 per km^2","code":"result=(240/0.6==400)"}
]